 # Let G be a group of order p^m where p is prime number and m is a positive integer. Show that G contains an element of order p. foass77W 2021-01-27 Answered
Let G be a group of order ${p}^{m}$ where p is prime number and m is a positive integer. Show that G contains an element of order p.
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Let G be a group of order ${p}^{m}$, where p is a prime number and m is a positive integer.
Then,
$|G|={p}^{m}=p{p}^{m-1}=pn,$
Where $n={p}^{m-1}$
Then, G is a finite group and the prime p divides the order of G, i.e.,
Proving the result by induction on n,
Let n=1
Then, ${p}^{m-1}=1$ and hence p=1.
Then, |G|=1
Thus, the group has only one element of order 1. The result holds for n=1
Next, suppose every group with pk elements contains an element of order p for every k Consider the class equation of G, $|G|=|Z\left(G\right)|+\sum _{a\ne Z\left(G\right)\left[G:C\left(a\right)\right]}$
If G=Z(G), then G is abelian and hence the result follows.
Suppose $G\notin Z\left(G\right)$
Then, for every a !in Z(G),
$C\left(a\right)\subset G$, and hence $|C\left(a\right)|.
If for some $a\notin Z\left(G\right),p|C\left(a\right)|$ then, by the induction hypothesis, there is an element a of order $p\in C\left(a\right)$ and hence $\in G$.
Otherwise, for every $a\notin Z\left(G\right),p\mid |C\left(a\right)|.$
Then, $p\mid \sum _{a\ne Z\left(G\right)\left[G:C\left(a\right)\right]\mid }$ and the class equation implies that $p|Z\left(G\right)|.$
Thus, it follows that Z(G) and hence G contains an element of order p.